Entire function

Could an entire function be determined by its zeros?

This is the core problem we would focus in this chapter.

Jensen's Formula

Mean value Theorem

Assume \(f\) is holomorphic on a disc \(D_R(z_0)\), and vanishes nowhere on the circle \(C_R\), then

\[ f(z_0)=\frac{1}{2\pi}\int_0^{2\pi} f(z_0+Re^{i\theta})d\theta. \]

Since taking a real part and integration could be exchanged, so

\[ \text{Re } f(z_0)=\frac{1}{2\pi}\int_0^{2\pi} \text{Re } f(z_0+Re^{i\theta})d\theta. \]

Proof

Rewrite the Cauchy's integral formula with \(\zeta=z_0+Re^{i\theta}\).

\(\square\)

Corollary

Assume a holomorphic function \(g\) on \(D_R(z)\) that vanishes nowhere in the closure of \(D_R(z_0)\) (not vanish on the integral circle), then

\[ \log |g(z_0)|=\frac{1}{2\pi}\int_0^{2\pi}\log |g(z_0+Re^{i\theta})|d\theta. \]

Proof

Just by Theorem for generating logarithm of a function, and we have \(g(z)=e^{h(z)}\), apply Mean value Theorem to \({\text{Re }h(z)}\). Notice \(\log |g(z)|={\text{Re }h(z)}\).

\(\square\)

Jensen's Formula

Let \(\Omega\) be an open set that contains the closure of a disc \(D_R\) and suppose \(f\) is holomorphic in \(\Omega\), \(f(0)\neq 0\) and \(f\) vanishes nowhere on the circle \(C_R\). If \(z_1,\cdots, z_n\) denote the zeros of \(f\) inside the disc (counted with multiplicities), then

\[ \log |f(0)|=\sum_{k=1}^N \log \left(\frac{|z_k|}{R}\right)+\frac{1}{2\pi}\int_0^{2\pi} \log|f(Re^{i\theta})|d\theta. \]

Proof

• Show that if \(f_1\) and \(f_2\) satisfies the above theorem, then \(f_1f_2\) also satisfies.

• Decompose \(f\) into

\[ f(z)=\prod_{k=1}^n(z-z_k) g(z) \]

where \(g(z)\) vanishes nowhere in the closure of \(D_R(z_0)\). Show that Jensen's Formula holds for function \(g\) and \(z-z_k\).

• For \(g(z)\), apply the above corollary with the same radius.

• For \(F(z)=z-w\), where \(|w|<|R|\).

We only need to prove that

\[ 0=\int_{0}^{2\pi} \log |e^{i\theta}-a|d\theta=\int_{0}^{2\pi} \log |1-ae^{i\theta}|d\theta \]

where \(a=w/R\). Define \(h(z)=1-az\), which vanishes nowhere in the closure of \(D_R(z_0)\). Apply the above corollary to it with integration path of unit circle and find that \(\log |h(0)|=0\).

\(\square\)

The Jensen's Formula connects the number of zeros of a holomorphic function \(f\) (in disc \(D_R\)) inside a circle. We define \(\mathscr{n} (r)\) the number of zeros of \(f\) inside the circle \(C_r\) where \(0<r<R\). Apparently, \(\mathscr{n}(r)\) is non-decreasing of \(r\).

Lemma: number of zeros expressed by Characteristic function

If \(z_1,\cdots,z_N\) are the zeros of \(f\) inside the disc \(D_R\), then

\[ \int_0^R \mathscr{n}(r)\frac{dr}{r}=\sum_{k=1}^N \log \left|\frac{R}{z_k}\right|. \]

Proof

Note the real logarithm and change the integral domain

\[ \begin{align*} \sum_{k=1}^N \log \left|\frac{R}{z_k}\right|&=\sum_{k=1}^N \log R - \log |z_k|\\ &=\sum_{k=1}^N \int_{|z_k|}^R \frac{dr}{r}\\ &=\sum_{k=1}^N \int_0^R 1_{\{r>z_k\}}\frac{dr}{r}\\ &=\int_0^R \sum_{k=1}^N 1_{\{r>z_k\}}\frac{dr}{r}\\ &=\int_0^R \mathscr{n}(r)\frac{dr}{r}. \end{align*} \]

\(\square\)

By the above lemma and Jensen's formula, we have

\[ \begin{align} \int_0^R \mathscr{n}(r)\frac{dr}{r}=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})d\theta - \log|f(0)|.\label{Jensen-number of zeros} \end{align} \]

for \(f(0)\neq 0\) and \(f\) does not vanish on the circle \(C_R\). Readers could check the following estimate for \(\mathscr{n}(r)\).

Function of finite order

Definition of finite order

Let \(f\) be an entire function. If there exists a positive number \(\rho\) and constants \(A,B>0\), such that

\[ |f(z)|\leq Ae^{B|z|^\rho},\quad \forall z\in \mathbb{C} \]

then \(f\) is called to have an order of growth \(\leq \rho\). Its order of growth id defined by

\[ \rho_f =\inf \rho. \]

Theorem

If \(f\) is an entire function that has an order of growth \(\leq \rho\), then

(i) \(\mathscr{n}(r)\leq Cr^\rho\) for some \(C>0\) and all sufficiently large \(r\).

(ii) If \(\{z_k\}_{k\geq 1}\) denotes the zeros of \(f\), with \(z_k\neq 0\), then for all \(s>\rho\), we have

\[ \sum_{k=1}^\infty \frac{1}{|z_k|^s}<\infty. \]

Proof

(i) By equation \(\ref{Jensen-number of zeros}\), if \(f(0)\neq 0\), we have

\[ \int_0^R\mathscr{n}(x)\frac{dx}{x}=\frac{1}{2\pi}\int_0^{2\pi} \log |f(Re^{i\theta})|d\theta - \log |f(0)| \]

If \(f(0)=0\), we let \(F(z)=\frac{f(z)}{z^l}\), where \(l\) is the order of zeros at \(z=0\), then \(F(0)\neq 0\), and we could still use the above formula.

The following is a common trick. For a fixed \(r\), let \(R=2r\), then we have

\[ \int_0^R\mathscr{n}(x)\frac{dx}{x}\geq \int_r^{2r}\mathscr{n}(x)\frac{dx}{x}\geq \mathscr{n}(r)\int_r^{2r}\frac{dx}{x}=\mathscr{n}(r)\log 2 . \]

Then on the other hand, since \(|f(z)|\leq A e^{B|z|^\rho}\), we have

\[ \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} \log |f(Re^{i\theta})|d\theta&\leq \frac{1}{2\pi}\int_0^{2\pi} \log |Ae^{BR^\rho}|d\theta=\log |A|+B2^\rho r^\rho\leq C' r^\rho. \end{align*} \]

Combining the above two inequations, we are done.

(ii) We use a bisection method like we used in proving the convergence of \(\sum_{k=1}^\infty \frac{1}{k^2}\).

\[ \begin{align*} \sum_{k=1}^\infty \frac{1}{|z_k|^s}&=\sum_{j=0}^\infty\left(\sum_{2^j\leq |z_k|< 2^{j+1}} \frac{1}{|z_k|^s}\right)\\ &\leq \sum_{j=0}^\infty\left(\sum_{2^j\leq |z_k|< 2^{j+1}} \frac{1}{|2^j|^s}\right)\\ &\leq \sum_{j=0}^\infty\left(\sum_{|z_k|< 2^{j+1}} \frac{1}{2^{js}}\right)\\ &\leq \sum_{j=0}^\infty\left(\mathscr{n}(2^{j+1})\frac{1}{2^{js}}\right)\\ &\leq \sum_{j=0}^\infty \left(C2^{(j+1)\rho}\frac{1}{2^{js}}\right)\\ &\leq C 2^\rho\sum_{j=0}^\infty 2^{(\rho-s)j}<\infty.\\ \end{align*} \]

Infinite Product

Now let us focus on constructing a function with given zeros \(\{z_k\}\), where \(\lim\limits_{k\rightarrow\infty}|z_k|=\infty\) (otherwise zeros would accumulate and cause the function to be zero identically). A naive guess is to define

\[ \begin{align} f(z)=\prod_{k=1}^\infty (z-z_k).\label{naive-version} \end{align} \]

Weierstrass gave a condition, and added an factor into the infinite product, for which to be convergent and no more zeros introduced.

Generalities

We first link the convergence of infinite products with that of series. In standard language, we focus on

\[ \prod_{n=1}^\infty (1+a_n). \]

Convergence of infinite product of standard form

If \(\sum_n|a_n|<\infty\), then the product

\[ \prod_{n=1}^\infty (1+a_n) \]

converges. Moreover, it converges to \(0\) iff one of its factors is \(0\).

Proof

Just use the logarithm to transfer the product into series. For \(1+a_n\neq 0\), we have

\[ \prod_{n=1}^N(1+a_n)=\prod_{n=1}^N e^{\log (1+a_n)}=\exp \left(\sum_{n=1}^N \log (1+a_n)\right) \]

Since exponential function is continuous, we only need to show the convergence of \(\sum\limits_{n=1}^N \log (1+a_n)\). Note that

\[ \log |1+a_n|\leq |\log (1+a_n)| \]

we could not use \(\log |1+a_n|\leq |a_n|\) to prove. Rather, we have to show that

\[ \begin{align} |\log (1+a_n)|\leq C|a_n|\label{tangent scaling} \end{align}\]

for some \(a_n\). Then the above infinite product converges. Actually, if we dig further, we find that

\[ \begin{align*} |\log(1+z)|&=\left|-\sum_{k=1}^\infty (-1)^k\frac{z^k}{k}\right|\\ &\leq\sum_{k=1}^\infty \frac{|z|^k}{k}\quad \text{triangle inequation}\\ &=|z|\sum_{k=0}^\infty \frac{|z|^k}{k+1}\\ &\leq |z|\sum_{k=0}^\infty |z|^k\\ \end{align*} \]

we try to let the rest series to converge, so just let \(|z|\leq \delta<1\) and we have \(|\log(1+z)|\leq C |z|\) for \(C=\frac{1}{1-\delta}\). In the above case, for large \(n\), we must have \(|a_n|<\delta\). So inequation \(\ref{tangent scaling}\) holds, if we just disregard the finitely many terms.

\(\square\)

More generally, we could consider the convergence of products of holomorphic functions. Note its convergence is also uniform, i.e. regardless of \(z\).

Products of holomorphic functions

Assume a sequence of holomorphic functions \(\{F_n\}\) on an open set \(\Omega\). If \(\exists c_n>0\) such that \(\sum_{n}c_n<\infty\) and

\[ |F_n-1|\leq c_n,\quad \forall n, \]

then

\[ \prod_{n=1}^\infty F_n(z) \]

converges uniformly in \(\Omega\) to a holomorphic function \(F(z)\). Moreover, if for each \(n\), \(F_n(z)\) does not vanish, then we have the logarithm derivative additivity holds

\[ \frac{F'(z)}{F(z)}=\sum_{n=1}^\infty \frac{F_n'(z)}{F_n(z)}. \]

Proof

The uniform convergence is apparent according to Weierstrass test. Check that \(a_n(z)=F_n(z)-1\),

\[ \begin{align*} \left|\prod_{n=1}^N F_n(z)\right|&\leq \exp\left|\sum_{n=1}^N \log F_n(z)\right|\\ &\leq \exp\left(\sum_{n=1}^N \left|\log F_n(z) \right|\right)\\ &\leq \exp \left(\sum_{n=1}^N \left|F_n(z)-1 \right|\right)\\ &\leq \exp \left(\sum_{n=1}^N |a_n|\right)\leq \exp \left(\sum_{n=1}^N c_n\right)<\infty,\quad \forall z\in \Omega. \end{align*} \]

Define

\[ G_N(z)=\prod_{n=1}^N f_n(z) \]

then \(G_N(z)\) converges uniformly to \(F(z)\) in \(\Omega\). So for an arbitrary compact subset \(K\subset \Omega\), by Sequence of holomorphic functions, \(G_N'(z)\) converges to \(F'(z)\) in \(K\). So \(\frac{G_N'}{G_N}\) converges to \(\frac{F'}{F}\) in \(K\). Expand \(\frac{G_N'}{G_N}\) by additicity of logarithm, i.e.

\[ \frac{G_N'(z)}{G_N(z)}=\sum_{n=1}^N \frac{F_n'(z)}{F_n(z)} \]

Let \(N\rightarrow \infty\) and we are done.

\(\square\)

Example. Show that

\[ \frac{\sin \pi z}{\pi}=z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right). \]

Weierstass Infinite Product

We fucos on adding the following factors into the vaive version of infinite product \(\ref{naive-version}\).

Canonical Factors

For each \(k\geq 0\), we define the Canonical factors to be

\[ E_0(z)=1-z, E_k(z)=(1-z)\exp\left(\sum_{n=1}^k z^n/n\right). \]

Lemma: order esstimate of canonical factors

If \(|z|<\frac{1}{2}\), then \(|1-E_k(z)|\leq c|z|^{k+1}\) for some constant \(c\).

Proof

Similar as we talked in Products of holomorphic functions. Using Taylor expansion of \(\log (1-z)\), we have

\[ \begin{align*} (1-z)\exp\left(\sum_{n=1}^k z^n/n\right)&=\exp\left(\log (1-z)+\sum_{n=1}^k z^n/n\right)\\ &=\exp\left(-\sum_{n=1}^\infty z^n/n+\sum_{n=1}^k z^n/n\right)\\ &=\exp\left(-\sum_{n=k+1}^\infty z^n/n\right) \end{align*} \]

Define \(w=\log (1-z)+\sum_{n=1}^k z^n/n\). So

\[ \begin{align*} \left|w\right|&=\left|\left(-\sum_{n=k+1}^\infty z^n/n\right)\right|\\ &\leq \left|-\sum_{n=k+1}^\infty z^n/n\right|\\ &\leq \left(\sum_{n=k+1}^\infty |z|^n/n\right)\\ &\leq \left(|z|^{k+1}\sum_{n=0}^\infty |z|^n\right)\\ &\leq {2|z|^{k+1}}\quad \text{using } |z|<\frac{1}{2}<1 \end{align*} \]

And apply Taylor's expansion to \(1-e^w\) again

\[ \begin{align*} |1-E_k(z)|&=|1-e^w|\\ &=\left|1-\sum_{n=0}^\infty w^n/n!\right|\\ &=\left|-\sum_{n=1}^\infty w^n/n!\right|\\ &\leq \sum_{n=1}^\infty |w|^n/n!\\ &\leq |w|\sum_{n=0}^\infty |w|^n/(n+1)!<e|w|,\quad \text{for } |w|<1.\\ &\leq 2e|z|^{k+1}. \end{align*} \]

where \(c=2e\).

\(\square\)

Weierstass Infinite Product

Given any sequence \(\{a_n\}\) of complex number with \(|a_n|\rightarrow \infty\) as \(n\rightarrow\infty\), there exists an entire function \(f\) that vanishes at all \(z=a_n\) and nowhere else. Any other such function is of the form \(f(z)e^{g(z)}\), where \(g(z)\) is an entire function.

Proof

Suppose \(f\) has an order \(m\) of zeros at \(z=0\), and other zeros \(\{z_n\}\), define Weierstrass infinite product

\[ f(z)=z^m \prod_{n=1}^\infty E_n(z/z_n). \]

We only need to show that it is convergent, which vanishes nowhere else. The Proof is a common trick, which partition the product into two separate parts. We only need to prove the convergence in open set \(\{|z|<R\}\) for each fixed \(R\).

\[ \prod_{n=1}^\infty E_n(z/z_n)=\prod_{|a_n|< 2R}E_n(z/z_n)\cdot \prod_{|a_n|\geq 2R} E_n(z/z_n). \]

For the former part of the righthand side, we have finite many terms, which converges, and has a zeros of the desired.

For the latter part of the righthand side, we have \(|z|/|a_n|<\frac{R}{2R}<\frac{1}{2}\), and by Lemma: order esstimate of canonical factors, \(|1-E_n(z/a_n)|\leq c|z/a_n|^{n+1}<c (1/2)^{n+1}\), by Products of holomorphic functions, it converges to a holomorphic function.

• For two functions that satisfies the above condition, denoted by \(f_1\), \(f_2\), then \(f_1/f_2\) has removable singularies at \(\{z_n\}\), so it is entire and vanish nowhere, by Theorem for generating logarithm of a function, \(\exists g(z)\) that is holomorphic on \(\mathbb{C}\)(thus entire), such that

\[ \frac{f_1}{f_2}=e^{g(z)}. \]

So \(f_1=f_2e^{g(z)}\).

Hadamard's factorization theorem

Hadamard gives a refinement of Weierstrass's Theorem about infinite products by fixing the degree of canonical factors and prove \(g(z)\) is a polynomial when \(f\) is of finite order of growth.

Hadamard's factorization theorem

Assume \(\{a_n\}\) are a sequence of points with \(\lim\limits_{n\rightarrow \infty}|a_n|=\infty\). For an entire function \(f\) with zeros \(\{a_n\}\) and order of growth \(\rho_0\), there exsits \(k\) such that \(k\leq \rho_0<k+1\), and

\[ f(z)=e^{g(z)}z^m\prod_{n=1}^\infty E_k(z/a_n) \]

where \(g(z)\) is a polynomial of degree less then \(k\) and \(m\) is the order of the zero of \(f\) at \(z=0\).